SOLUTION: Suppose you have an urn containing 7 red and 3 blue balls. You draw three balls at random. On each draw, if the ball is red you set it aside and if the ball is blue you put it back

Question 1103609: Suppose you have an urn containing 7 red and 3 blue balls. You draw three balls at random. On each draw, if the ball is red you set it aside and if the ball is blue you put it back in the urn. What is the probability that the third draw is blue? (If you get a blue ball it counts as a draw even though you put it back in the urn)
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!
Suppose you have an urn containing 7 red and 3 blue balls.
You draw three balls at random. On each draw, if the ball is red
you set it aside and if the ball is blue you put it back in the
urn. What is the probability that the third draw is blue? (If
you get a blue ball it counts as a draw even though you put it
back in the urn)

The possible successful cases are

case 1. RRB
case 2. RBB
case 3. BRB
case 4. BBB

-----

case 1. RRB

The probability of drawing a red first is 7/10
Then you set it aside and the urn now contains 6 reds and 3 blues
The probability of drawing a red second is 6/9 or 2/3
Then you set it aside and the urn now contains 5 reds and 3 blues
The probability of drawing a blue third is 3/8

So the probability for case 1 is (7/10)(2/3)(3/8) = 7/40

case 2. RBB

The probability of drawing a red first is 7/10
Then you set it aside and the urn now contains 6 reds and 3 blues
The probability of drawing a blue second is 3/9 or 1/3
Then you put it back and the urn still contains 6 reds and 3 blues
The probability of drawing a blue third is 3/9 or 1/3

So the probability for case 2 is (7/10)(1/3)(1/3) = 7/90

case 3. BRB

The probability of drawing a blue first is 3/10
Then you put it back and the urn still contains 7 reds and 3 blues
The probability of drawing a red second is 7/10 
Then you set it aside and the urn now contains 6 reds and 3 blues
The probability of drawing a blue third is 3/9 or 1/3

So the probability for case 2 is (3/10)(7/10)(1/3) = 7/100


case 4. BBB

The probability of drawing a blue first is 3/10
Then you put it back and the urn still contains 7 reds and 3 blues
The probability of drawing a blue second is 3/10 
Then you put it back and the urn still contains 7 reds and 3 blues
The probability of drawing a blue third is 3/10 

So the probability for case 4 is (3/10)(3/10)(3/10) = 27/1000

----

So the final probability is the sum of those 4 probabilities:

7/40 + 7/90 + 7/100 + 27/1000 = 787/2250

That's about 0.34977777...

Edwin

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